Can a Baseball Thrown at 25 m/s Hit a Window 13 Meters High and 50 Meters Away?

AI Thread Summary
A student is trying to determine if a baseball thrown at 25 m/s can hit a window 13 meters high and 50 meters away. The discussion emphasizes that the angle of projection affects the maximum height achievable by the baseball. While a 45-degree angle maximizes horizontal distance, the fixed range of 50 meters means a different angle may be needed to reach the required height. Participants suggest using equations for horizontal and vertical motion to find the optimal angle for hitting the window. Ultimately, the solution involves calculating the angle that maximizes the height when the baseball reaches 50 meters horizontally.
descartes75
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Hi,
There has to be a simple way of solving this and I am missing it:

A student throws a baseball at a maximum speed of 25 m/s towards a window 50 meter away (horizontally) and 13 meters high. Can he hit the window? (assume the base of the window starts at 13 meters off the ground).
What is the maximum hight he can reach from this distance??


Since the Angle is now known and depends on the maximum hight, I am having trouble solving this. Any help?
 
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The angle is may not be known but it may be useful to know that 45 degrees is the angle that will give a projectile maximum range in this case, so if the ball can not make it at 45 degrees, then it can not make it period.
 
treat horizontal velocity and vertical velocity separetely,
the horizontal velocity is: 25 * cos(throw_angle)
the vertical velocity is: 25 * sin(throw_angle)

you will want to ignore air resistance, so the horizotal velocity does not change but the vertical velocity does change because of gravity, per time unit it will change 1 g.
so the horizontal velocity will be:
25 * cos(angle)
and the vertical velocity will be:
25 * sin(angle) - gt

can you see why an angle of 45 degrees gets the ball the furthest distance?
 
Thanks Guys.
But isn't 45 degree for maximum range (HOrizontal distance) ??
In this case the range is fixed (50) so intuitively the angle doesn't have to be 45 degrees.

You can imagine standing in front of a tall building with a ball and aiming for a high window. You won't ignore intuition and throw at a 45 degree angle (and hit the door).
 
descartes75 said:
Since the Angle is now known and depends on the maximum hight, I am having trouble solving this. Any help?
You'll have to figure out what initial angle gives the maximum height at X = 50m. Start by writing equations for X and Y as a function of time. Then write an expression for Y at the moment that X = 50m. It will be a function of \theta. Find the value of \theta that maximizes this function. (Take a derivative.)
 
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